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Let's say $A$ is an orthogonal $2\times2$ matrix over $\bf C$ and not diagonalizable over $\bf C$. Why then the determinant of $A$ must be $1$?

I guess I'm missing something easy...

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Do you have an example of an orthogonal matrix not diagonalizable over the complex numbers? –  Gerry Myerson Feb 2 '13 at 23:46
    
No I don't why? –  baaa12 Feb 2 '13 at 23:47
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Why would you ask, if you didn't have an example, or some other reason for believing it to be true? Why do you believe it to be true? –  Gerry Myerson Feb 2 '13 at 23:52
    
This is a multiple choices question when I need to mark to correct answer and according to my teacher this is a correct answer –  baaa12 Feb 2 '13 at 23:56
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Well, it's correct in the sense that the hypothesis is false so any conclusion is true. But it would be just as true to say that if $A$ is orthogonal and not diagonalizable over the complex numbers then its determinant must be $17$. I wonder whether the question is not supposed to be about diagonalizability over the reals? –  Gerry Myerson Feb 3 '13 at 0:41
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3 Answers 3

Orthogonal matrices are normal, and all normal matrices are diagonalizable over the complex numbers.

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Was trying to prove it and it got so messy. This is so amazingly clean! –  Inquest Feb 3 '13 at 0:04
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This is false, e.g. $$\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$$

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Forgot to mention matrix is orthogonal –  baaa12 Feb 2 '13 at 23:43
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Edit: It is well known that every real orthogonal matrix is diagonalizable over $\mathbb{C}$. Actually, every $2\times 2$ complex orthogonal matrix is also diagonalizable over $\mathbb{C}$. See theorem 1.2.3 of this thesis, for instance. So, your assertion can be viewed as a vacuous truth.

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